I am puzzled by this computation: https://www.dsprelated.com/showarticle/754.php (c.f. quote)

Raising $ i $ to integer powers results in traversing the unitcircle in the same number of quarter turns. The next question that arises naturally is if the pattern also applies to fractional values as well. Consider $ \sqrt{i} $. Suppose it is $ a + b \cdot i $. $$ ( a + b \cdot i )^{2} = i \\ a^{2} + 2ab \cdot i + b^{2} \cdot i^{2} = i $$ $$ ( a^{2} - b^{2} ) + ( 2ab ) \cdot i = 0 + 1 \cdot i $$ In order for two complex numbers to be equal, both the real parts and the imaginary parts must be equal. $$ a^{2} - b^{2} = 0 \text{ and } 2ab = 1 $$ $$ a = ± b \text{ and } a = { 1 \over {2b} } $$ Solving for a,b being real yields two solutions: $$ (a,b) = \left( { \sqrt{2} \over 2 }, { \sqrt{2} \over 2 } \right) \text{ or } \left( - { \sqrt{2} \over 2 }, - { \sqrt{2} \over 2 } \right) $$

For me, it does not make sense to say that $a=\pm b$ AND $2ab=1$, this is just a contradiction... either $a=\pm b$ OR $2ab=1$ therefore I would say that $(a+bi)^2 \neq i$ .... How can they say that it is? What is the logic behind that...?


2 Answers 2


That's my article, I'm glad you are reading it.

A fundamental rule for two complex numbers (corresponding to a point on a plane) is that they are only equal if both the real and imaginary parts (corresponding to x,y coordinates) are equal. Therefore the complex equation

$$ (a^2 - b^2) + 2abi = i = 0 + i $$

is equivalent to saying the point $(a^2 - b^2,2ab)$ is the same point as $(0, 1)$

The real parts must be equal AND the imaginary parts must be equal.

Graphically, you have the intersection of a hyperbola intersecting crossed lines going through the origin. Substitute $x$ and $y$ for $a$ and $b$ if that helps you see it.

The important concept is that an intersection point is halfway along the unit circle, and the other one is directly opposite (the negative of it.) This is true for taking the square root of any number on the unit circle. One square root will be halfway and the other the negative of it.

And that is part of the exponential nature of the unit circle. Radians is just one possible scale. Quarter turns is another, whole turns another.

Expanding on this, the solutions to $1^{1/N}$ are $N$ evenly spaced points around the circle. They are called the Roots of Unity. You may know the first root better as $e^{i2\pi/N}$ which is right there in the DFT definition:

$$ X[k] = \sum_{n=0}^{N-1} x[n] \left(e^{i2\pi/N}\right)^{-nk} $$

$$ a^2 - b^2 = 0 $$

$$ a^2 = b^2 $$

$$ a = \pm b $$


$$ 2ab = 1 $$

For $a=b$

$$ 2a^2 = 1 $$

$$ a = \pm \frac{\sqrt{2}}{2} = b $$

For $a=-b$

$$ -2a^2 = 1 $$

No real solutions.

  • $\begingroup$ Thanks, but shouldn't you rather say (a^2-b^2) and 2a'b' (primes or anything else) or another name for the ab variables because one would want to consider that "a" is the same "a" in (a^2-b^2) and 2ab ... do you see my problem? (If i understand correctly the a's and b's in the real part and in the imag part are not nessassarily the same) $\endgroup$
    – SheppLogan
    Jul 20, 2019 at 12:35
  • 1
    $\begingroup$ @Machupicchu No, I don't see your problem, they are the same $a$ and $b$ from the posited value $a+bi$. $\endgroup$ Jul 20, 2019 at 12:37
  • $\begingroup$ but don't you have a 2 equations 2 unknown system of equations : (a^2 - b^2) = 0, and, 2ab=1 ? $\endgroup$
    – SheppLogan
    Jul 20, 2019 at 12:38
  • 1
    $\begingroup$ Precisely. What real values of $a$ and $b$ satisfy those criteria. I suggest you get out some graph paper, draw a unit circle, then graph: $$ y = x $$ $$ y = -x $$ $$ y = \frac{1}{2x} $$ Then notice where the interestion points lie. Note too that $x^2+y^2=1$ at your intersection points. $\endgroup$ Jul 20, 2019 at 12:42
  • $\begingroup$ Oh by solving it I get : $a=1/2b$ then , substituting in $(a^2-b^2)=0$ I get: $ ((1/(4b^2))^2 -b^2)=0 => b^4 = 1/4 => b=(1/4)^{1/4} = \sqrt(2)/2$ Which is your solution ...? $\endgroup$
    – SheppLogan
    Jul 20, 2019 at 12:45

You have two valid sets of equations, where $a$ and $b$ are real:

  • $a=b$ and $2ab=1$
  • or $a=-b$ and $2ab=1$

The second one a no solution. The first has two solutions (so you have two solutions in total, as written), that are $\frac{\sqrt{2}}{2}(1+i)$ and $-\frac{\sqrt{2}}{2}(1+i)$. In other words, the "positive one-eighth turn", and "the negative one-eighth turn" both yield $i$ (the quarter turn) when raised to the square. This is very similar to real numbers: given a real value $r$, both $r^2 $ and $(-r)^2 $ give the same results.

By let's go beyond that. A positive number is a 0-degree turn. A negative number is a 180-degree turn. Both raised to the square are positive, and thus yield some 0-degree turn at the end. Although complex numbers seem historically less intuitive than real ones, they are perfectly consistent with them (would they have failed to do so, they would not have survived). Moreover, from a modern perspective, I'd say that complex numbers are the ones which really "exist", and we humans can merely see their shadows (real numbers), albeit with some clever visualizations for us blind, like the Heyser corkscrew/spiral:

Heyser corkscrew spiral

  • $\begingroup$ Thanks, I didn't see this until after I appended the solution. $\endgroup$ Jul 20, 2019 at 13:03
  • $\begingroup$ I tweaked your answer a bit. Notice that the definition of the DFT can now be seen as either the weighted sum of Roots of Unity (any Root of Unity raised to an integer power is another of that set) or the sum of rotated signal values. My followup article dsprelated.com/showarticle/768.php explores the first with a $1/N$ normalized DFT making it a weighted average. The subsequent articles develop novel equations about pure tones in the DFT completely unreliant on any reference to the continuous case. I.e. it is fundamental (and much simpler). Hope you read them if you haven't. $\endgroup$ Jul 20, 2019 at 13:33

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